Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 901: 28

Answer

$\frac{4}{\pi^2}$

Work Step by Step

$ V=\int^1_0 \int^{1-x}_0 \int^{cos(\pi x/2)}_0$ dz dy dx =$\int^1_0 \int^{1-x}_0 cos(\frac{\pi x}{2}dy)$ dx =$\int^1_0 (cos\frac{\pi x}{2})(1-x)dx $ =$\int^1_0 cos (\frac{\pi x}{2})dx-\int^1_0 x \cos (\frac{\pi x}{2}) dx $ =$[\frac{2}{\pi} sin \frac{\pi x}{2}]^1_0 -\frac{4}{\pi^2} \int^{\pi/2}_0 u \cos u du $ =$\frac{2}{\pi}-\frac{4}{\pi^2}[\cos u+u\sin u]^{\pi/2}_0$ =$\frac{2}{\pi}-\frac{4}{\pi^2}(\frac{\pi}{2}-1)$ =$\frac{4}{\pi^2}$
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